5-121
5-124 Prob. 5-123 is reconsidered. The temperature at the top corner as a function of heating time is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
T_i=140 [C]
k=15 [W/m-C]
alpha=3.2E-6 [m^2/s]
e_dot=2E7 [W/m^3]
T_bottom=140 [C]
T_infinity=25 [C]
h=80 [W/m^2-C]
q_dot_L=8000 [W/m^2]
DELTAx=0.015 [m]
DELTAy=0.015 [m]
time=120 [s]
“ANALYSIS”
l=DELTAx
DELTAt=15 [s]
tau=(alpha*DELTAt)/l^2
RhoC=k/alpha “RhoC=rho*C”
“The technique is to store the temperatures in the parametric table and recover them (as old temperatures)
using the variable ROW. The first row contains the initial values so Solve Table must begin at row 2.
Use the DUPLICATE statement to reduce the number of equations that need to be typed. Column 1
contains the time, column 2 the value of T[1], column 3, the value of T[2], etc., and column 10 the Row.”
Time=TableValue(‘Table 1′,Row-1,#Time)+DELTAt
Duplicate i=1,8
T_old[i]=TableValue(‘Table 1′,Row-1,#T[i])
end
“Using the explicit finite difference approach, the eight equations for the eight unknown temperatures are
determined to be”
q_dot_L*l/2+h*l/2*(T_infinity-T_old[1])+k*l/2*(T_old[2]-T_old[1])/l+k*l/2*(T_old[4]-
T_old[1])/l+e_dot*l^2/4=RhoC*l^2/4*(T[1]-T_old[1])/DELTAt “Node 1″
h*l*(T_infinity-T_old[2])+k*l/2*(T_old[1]-T_old[2])/l+k*l/2*(T_old[3]-T_old[2])/l+k*l*(T_old[5]-
T_old[2])/l+e_dot*l^2/2=RhoC*l^2/2*(T[2]-T_old[2])/DELTAt “Node 2″
h*l*(T_infinity-T_old[3])+k*l/2*(T_old[2]-T_old[3])/l+k*l/2*(T_old[6]-T_old[3])/l+e_dot*l^2/4=RhoC*l^2/4*(T[3]-
T_old[3])/DELTAt “Node 3″
q_dot_L*l+k*l/2*(T_old[1]-T_old[4])/l+k*l/2*(T_bottom-T_old[4])/l+k*l*(T_old[5]-
T_old[4])/l+e_dot*l^2/2=RhoC*l^2/2*(T[4]-T_old[4])/DELTAt “Node 4″
T[5]=(1-4*tau)*T_old[5]+tau*(T_old[2]+T_old[4]+T_old[6]+T_bottom+e_dot*l^2/k) “Node 5”
h*l*(T_infinity-T_old[6])+k*l/2*(T_old[3]-T_old[6])/l+k*l*(T_old[5]-T_old[6])/l+k*l*(T_bottom–
T_old[6])/l+k*l/2*(T_old[7]-T_old[6])/l+e_dot*3/4*l^2=RhoC*3/4*l^2*(T[6]-T_old[6])/DELTAt “Node 6″
h*l*(T_infinity-T_old[7])+k*l/2*(T_old[6]-T_old[7])/l+k*l/2*(T_old[8]-T_old[7])/l+k*l*(T_bottom-
T_old[7])/l+e_dot*l^2/2=RhoC*l^2/2*(T[7]-T_old[7])/DELTAt “Node 7″
h*l/2*(T_infinity-T_old[8])+k*l/2*(T_old[7]-T_old[8])/l+k*l/2*(T_bottom-
T_old[8])/l+e_dot*l^2/4=RhoC*l^2/4*(T[8]-T_old[8])/DELTAt “Node 8″
5-122
Time
[s]
T1
[C]
T2
[C]
T3
[C]
T4
[C]
T5
[C]
T6
[C]
T7
[C]
T8
[C]
Row
0
140
140
140
140
140
140
140
140
1
15
203.5
200.1
196.1
207.4
204
201.4
200.1
200.1
2
30
265
259.7
252.4
258.2
253.7
243.7
232.7
232.5
3
45
319
312.7
300.3
299.9
293.5
275.7
252.4
250.1
4
60
365.5
357.4
340.3
334.6
326.4
300.7
265.2
260.4
5
75
404.6
394.9
373.2
363.6
353.5
320.6
274.1
267
6
90
437.4
426.1
400.3
387.8
375.9
336.7
280.8
271.6
7
105
464.7
451.9
422.5
407.9
394.5
349.9
286
275
8
120
487.4
473.3
440.9
424.5
409.8
360.7
290.1
277.5
9
135
506.2
491
456.1
438.4
422.5
369.6
293.4
279.6
10
…
…
…
…
…
…
…
…
…
…
…
…
…
…
…
…
…
…
…
…
1650
596.3
575.7
528.5
504.6
483.1
411.9
308.8
288.9
111
1665
596.3
575.7
528.5
504.6
483.1
411.9
308.8
288.9
112
1680
596.3
575.7
528.5
504.6
483.1
411.9
308.8
288.9
113
1695
596.3
575.7
528.5
504.6
483.1
411.9
308.8
288.9
114
1710
596.3
575.7
528.5
504.6
483.1
411.9
308.8
288.9
115
1725
596.3
575.7
528.5
504.6
483.1
411.9
308.8
288.9
116
1740
596.3
575.7
528.5
504.6
483.1
411.9
308.8
288.9
117
1755
596.3
575.7
528.5
504.6
483.1
411.9
308.8
288.9
118
1770
596.3
575.7
528.5
504.6
483.1
411.9
308.8
288.9
119
1785
596.3
575.7
528.5
504.6
483.1
411.9
308.8
288.9
120
0 200 400 600 800 1000 1200 1400 1600 1800
100
150
200
250
300
350
400
450
500
550
Time [s]
T3 [C]
5-123
5-125 A long solid bar is subjected to transient two-dimensional heat transfer. The centerline temperature of the bar after 20
min and after steady conditions are established are to be determined.
Assumptions 1 Heat transfer through the body is given to be transient and two-dimensional. 2 Heat is generated uniformly in
the body. 3 The heat transfer coefficient also includes the radiation effects.
Properties The conductivity and diffusivity are given to be k = 28 W/m°C and
/sm 1012 26−
=
.
Analysis The nodal spacing is given to be x=x=l=0.1 m. The explicit finite
difference equations are determined on the basis of the energy balance for the
transient case expressed as
t
TT
cEQ
i
m
i
m
p
i
−
=+
+
1
element
i
element
sides All
V
• • •
• • •
h, T
7 8 9
h, T
• • •
4 5 6
1 2 3
h, T
e
5-124
5-126 ASTM A479 904L bar is submerged is hot fluid at T∞ = 300°C and h = 288 W/m2·K. The bar has an initial
temperature of 20°C. If the bar is submerged in the hot liquid for 7 minutes, would it be in compliance with the ASME code?
How long will it take for the bar to reach the maximum use temperature?
Assumptions1 The bars have square cross
section with infinite length. 2 Thermal properties
are constant. 3 Radiation effects are negligible. 4
The convection heat transfer coefficient is
constant. 5 Transient heat conduction is two–
dimensional.
Properties The thermal properties given are cp =
500 J/kg·K, k = 12 W/m·K, and ρ = 7900 kg/m3.
Analysis There is symmetry about the vertical,
horizontal, and diagonal lines passing through
the center. The symmetry lines are treated as
insulation boundary. Therefore, T1 = T3 = T7 = T9
and T2 = T4 = T6 = T8. The three unknown nodal
𝑘
4(1+ℎ∆𝑥 𝑘
⁄ )or ∆𝑡 ≤ ∆𝑥2
4𝛼(1+ℎ∆𝑥 𝑘
⁄ ) =32.15 s ⟹ ∆𝑡 = 10 s
where,
𝑐𝑝=500 J/kg∙K, 𝑘 = 12 W/m∙K, 𝜌 = 7900 kg/m3, ∆𝑥 = 2.5 cm
5-125
The numerical results can be solved iteratively with the initial nodal temperatures, Ti = 20°C, and time step, Δt = 10 s. The
following table tabulates the iterated nodal temperatures at every minute:
t [s]
T1 [°C]
T2 [°C]
T5 [°C]
0
20.0
20.0
20.0
60
142.6
102.9
53.5
120
191.8
157.0
111.1
180
221.7
195.1
159.6
240
242.6
222.9
196.4
300
257.8
243.2
223.7
360
268.9
258.2
243.8
420
277.1
269.2
258.7
480
283.1
277.4
269.6
The nodal temperatures, as a function of t, are plotted in the following figure:
0
50
100
150
200
250
300
0100 200 300 400 500
T [°C]
t [s]
T1
T2
5-127
The upper limit of the time step t is determined from the stability criteria that requires the coefficient of
i
m
T
in the
1+i
m
T
expression (the primary coefficient) be greater than or equal to zero for all nodes. The smallest primary coefficient in the 9
equations above is the coefficient of
i
T6
in the
1
6
+i
T
expression since it is exposed to most convection per unit volume (this
can be verified). The equation for node 6 can be rearranged as
+
+
+
+
+
+
−=
+
2
5
2
93
06
22
1
62
11
21
x
T
y
TT
T
xk
h
tT
xy
xk
h
tT
iii
o
i
o
i
Therefore, the stability criteria for this problem can be expressed as
+
+
→
+
+
−
22
22 11
2
1
t 0
11
21
xy
xk
h
xy
xk
h
t
o
o
Substituting the given quantities, the maximum allowable value of the time step is determined to be
or,
s 7.4
m) 01.0(
1
)m 002.0(
1
m) 002.0)(C W/m84.0(
C W/m20
)/m 1039.0(2
1
t
22
2
26
=
++
s
Therefore, any time step less than 4.8 s can be used to solve this problem. For convenience, we choose the time step to be t
= 4 s. Then the temperature distribution throughout the glass 15 min after the strip heaters are turned on and when steady
conditions are reached are determined to be
15 min: T1 = 10.2C, T2 = 10.2C, T3 = 9.8C, T4 = 16.0C, T5 = 15.9C,
5-128
5-128 The formation of fog on the glass surfaces of a car is to be prevented by attaching electric resistance heaters to the
inner surfaces. The temperature distribution throughout the glass 15 min after the strip heaters are turned on and also when
steady conditions are reached are to be determined using the implicit method with a time step of t = 1 min.
Assumptions 1 Heat transfer through the glass is given to be transient
and two-dimensional. 2 Thermal conductivity is constant. 3 There is
heat generation only at the inner surface, which will be treated as
prescribed heat flux.
Properties The conductivity and diffusivity are given to be k = 0.84
W/m°C and
/sm 1039.0 26−
=
.
Analysis The nodal spacing is given to be x = 0.2 cm and y = 1 cm.
The implicit finite difference equations are determined on the basis of
the energy balance for the transient case expressed as
t
TT
cEQ
i
m
i
m
p
i
−
=+
+
+
1
element
1+i
elementgen,
sides All
1
V
We consider only 9 nodes because of symmetry. Note that we do not
have a square mesh in this case, and thus we will have to rely on
Thermal
symmetry line
• • •
Outer
surface
1 2 3
Glass
Inner
surface
• • •
• • •
4 5 6
7 8 9
Thermal
symmetry line
5-129
5-129 Square cross section geometry is subjected to four different boundary conditions. The transient temperature
distribution within the geometry is to be determined after 15 seconds using the explicit finite difference formulation.
Assumptions 1 Two-dimensional transient heat conduction without internal heat generation. 2 Thermal properties are
constant.
Properties Thermal conductivity is given as k=20 W/m·K and the thermal diffusivity value is = 6.694 10-6 m2/s.
Analysis: (a) There are 4 internal nodes (node 2, 3, 6
and 7) and 4 boundary nodes (node 1, 4, 5 and 8). Thus
we need to have 8 equations for 8 unknown
temperatures. Finite difference equations for internal
5-130
( )
)s(69.3
)Km/W(20)m(01.0)Km/W(452)s/m(10694.62
)m(01.0
t226
22
+
−
For convenience let’s select a time step of
st 3=
.This gives a mesh Fourier number of
20082.0
2=
=y
t
(b) The temperature distribution is obtained by running the following EES code.
“Given data”
k = 20 [W/mC] ” Thermal conductivity”
h = 45 [W/m^2C] “ Convective heat transfer coefficient”
T_infi = 20 [C] ” Convective environment temperature”
q_dot = 1000 [W/m^2] ” Heat flux at left boundary”
DELTAt = 3 [s] “Time step”
DELTAy = 0.01 [m] ” Mesh size”
tau = 0.20082 “Mesh Fourier number”
“The technique is to store the temperatures in the parametric table and recover them (as old temperatures) using
the variable ROW. The first row contains the initial values so Solve Table must begin at row 2. Use the
DUPLICATE statement to reduce the number of equations that need to be typed. Column 1 contains the time,
column 2 the value of T[1], column 3, the value of T[2], etc., and column 9 the Row.To start the Solve Table at 2
go to ‘Calculate’ and select ‘Solve table’ (or hit F3) and make the ‘First Run Number’ as 2. The initial
temperatures and the initial time ‘0’ can be set manually in the parametric table”
Row = TableRun#
Time = TableValue(‘Table 1′,Row-1,#Time)+DELTAt
Duplicate i=1,8
T_old[i] = TableValue(‘Table 1′,Row-1,#T[i])
End
“Finite difference Equation”
“Node 1” T[1]=(1-4*tau)*T_old[1]+tau*(100+T_old[5]+2*T_old[2]+2*q_dot*DELTAy/k)
“Node 2” T[2] = (1-4*tau)*T_old[2]+tau*(100+T_old[6]+T_old[1]+T_old[3])
“Node 3” T[3] = (1-4*tau)*T_old[3]+tau*(100+T_old[2]+T_old[4]+T_old[7])
“Node4″T[4] = (1-4*tau-2*tau*h*DELTAy/k)*T_old[4]+tau*(100+T_old[8]+2*T_old[3]+2*h*DELTAy*T_infi/k)
“Node 5” T[5] = (1-4*tau)*T_old[5]+tau*(300+T_old[1]+2*T_old[6]+2*q_dot*DELTAy/k)
“Node 6” T[6] = (1-4*tau)*T_old[6]+tau*(300+T_old[5]+T_old[2]+T_old[7])
“Node 7” T[7] = (1-4*tau)*T_old[7]+tau*(300+T_old[6]+T_old[3]+T_old[8])
“Node8″T[8] = (1-4*tau-tau*2*h*DELTAy/k)*T_old[8]+tau*(300+T_old[4]+2*T_old[7]+2*h*DELTAy*T_infi/k)
Temperature distribution at different time steps.
Nodal temperature, oC
Time (s)
T1
T2
T3
T4
T5
T6
T7
T8
0
200.0
200.0
200.0
200.0
200.0
200.0
200.0
200.0
3
180.0
179.8
179.8
178.2
220.1
219.9
219.9
218.3
6
172.0
171.7
171.4
169.6
228.1
227.8
227.5
225.4
9
168.8
168.4
167.9
166.1
231.2
230.9
230.3
228.0
12
167.4
167.1
166.4
164.5
232.4
232.0
231.2
228.9
15
166.9
166.5
165.7
163.8
232.8
232.4
231.5
229.1
5-131
5-130 A copper alloy with known thermal properties and initially heated to 800oC is subjected to quenching in water with top
surface exposed to air environment. The temperature distribution in the copper alloy after 10 min is to be determined using
explicit finite difference formulation.
Assumptions 1 Two-dimensional transient heat transfer without internal heat generation. 2 Constant thermal properties 3
The top surface of the copper alloy is always in contact with the air environment.
Properties Thermal conductivity is given as
k = 120 W/m·K and the thermal diffusivity value
is = 3.91 10-6 m2/s.
Analysis The nodes 1, 2 and 3 are exposed to air
environment while nodes 4, 7, 8 and 9 are in
t
y
x
x
PROPRIETARY MATERIAL. © 2020 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
5-132
++++
−−=
+T
k
lh
TTTT
k
lh
Tiiiii 2
9758
2
1
822241
Node 9:
( ) ( ) ( ) ( )
t
TT
xc
y
y
TT
xk
x
TT
y
kTTxh
ii
p
iiii
i
−
=
−
+
−
+−
+
9
1
99698
92 22
2
+++
−−=
+T
k
lh
TTT
k
lh
Tiiii 2
869
2
1
9222241
The unknown nodal temperatures are found by running the following EES program.
“Given”
l = 0.1 [m] “mesh size”
alpha = 3.91e-6 “Thermal diffusivity [m^2/s]”
DELTAt = 10 [s] “Time step”
k = 120 [W/moC] “Thermal conductivity”
h_1 = 10 [W/m^2oC] “Convective heat transfer coefficient of air”
h_2 = 100 [W/m^2oC] “Convective heat transfer coefficient of water“
T_infi = 15 [C] ” Ambient temperature”
tau = (alpha*DELTAt)/l^2 “Mesh Fourier number”
“The technique is to store the temperatures in the parametric table and recover them (as old temperatures) using
the variable ROW. The first row contains the initial values so Solve Table must begin at row 2. Use the
DUPLICATE statement to reduce the number of equations that need to be typed. Column 1 contains the time,
column 2 the value of T[1], column 3, the value of T[2], etc., and column 9 the Row.
To start the Solve Table at 2 go to ‘Calculate’ and select ‘Solve table’ (or hit F3) and make the ‘First Run
Number’ as 2. The initial temperatures and the initial time ‘0’ can be set manually in the parametric table”
Row = TableRun#
Time = TableValue(‘Table 1′,Row-1,#Time)+DELTAt
Duplicate i=1,9
T_old[i] = TableValue(‘Table 1′,Row-1,#T[i])
end
“Finite difference formulation”
“Node 1” T[1] = (1-4*tau-2*tau*(h_1+h_2)*l/k)*T_old[1]+2*tau*(T_old[4]+T_old[2]+(h_1+h_2)*l*T_infi/k)
“Node 2” T[2] = (1-4*tau-2*tau*h_1*l/k)*T_old[2]+tau*(T_old[1]+T_old[3]+2*T_old[5]+2*h_1*l*T_infi/k)
“Node 3” T[3] = (1-4*tau-2*tau*h_1*l/k)*T_old[3]+tau*(2*T_old[2]+2*T_old[6]+2*h_1*l*T_infi/k)
“Node 4” T[4] = (1-4*tau-2*tau*h_2*l/k)*T_old[4] + tau*(T_old[1]+T_old[7]+2*T_old[5]+2*h_2*l*T_infi/k)
“Node 5” T[5] = (1-4*tau)*T_old[5]+tau*(T_old[2]+T_old[4]+T_old[6]+T_old[8])
“Node 6” T[6] = (1-4*tau)*T_old[6]+tau*(2*T_old[5]+T_old[3]+T_old[9])
“Node 7” T[7] = (1-4*tau-4*tau*h_2*l/k)*T_old[7]+2*tau*(T_old[4]+T_old[8]+2*h_2*l*T_infi/k)
“Node 8” T[8] = (1-4*tau-2*tau*h_2*l/k)*T_old[8]+tau*(T_old[7]+T_old[9]+2*T_old[5]+2*h_2*l*T_infi/k)
“Node 9” T[9] = (1-4*tau-2*tau*h_2*l/k)*T_old[9]+tau*(2*T_old[8]+2*T_old[6]+2*h_2*l*T_infi/k)
Temperature distribution in the copper alloy block after 10 min is as follows,
PROPRIETARY MATERIAL. © 2020 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
5-131 A ceramic strip at a specified initial temperature is exposed to convective environment at top and constant temperature
boundary conditions at its two sides. Using implicit finite difference formulation, the temperature distribution in the ceramic
strip is to be determined after 12 seconds.
Assumptions 1 Two-dimensional transient heat conduction without heat generation. 2 Thermal properties of the ceramic strip
stays constant.
Properties Thermal properties of ceramic strip are given as: k = 3 W/m·K, = 1600 kg/m3 and cp = 800 J/kg·K.
Analysis The finite difference equation at each
node is developed by doing an energy balance
assuming all heat transfer is to the control
volume of the node under consideration. Nodes
1, 2 and 3 are the boundary nodes exposed to the
“Given data”
k = 3 [W/mC] “Thermal conductivity of ceramic strip”
h = 200 [W/m^2C] “Convective heat transfer coefficient”
T_infi = 50 [C] ” Ambient temperature”
DELTAx = 0.01 [m] “Mesh size in x direction”
DELTAy = 0.01 [m] “Mesh size in y direction”
DELTAt = 2 [s] “Time step”
rho = 1600 [kg/m^3] “Density of ceramic strip”
c = 800 [J/kgK] ” Specific heat capacity of ceramic strip”
alpha = k/(rho*c) “Thermal diffusivity of ceramic strip [m^2/s]”
tau = DELTAx^2/(alpha*DELTAt)“Mesh Fourier number”
“The technique is to store the temperatures in the parametric table and recover them (as old temperatures) using
the variable ROW. The first row contains the initial values so Solve Table must begin at row 2. Use the
DUPLICATE statement to reduce the number of equations that need to be typed. Column 1 contains the time,
column 2 the value of T[1], column 3, the value of T[2], etc., and column 9 the Row.
To start the Solve Table at 2 go to ‘Calculate’ and select ‘Solve table’ (or hit F3) and make the ‘First Run
Number’ as 2. The initial temperatures and the initial time ‘0’ can be set manually in the parametric table”
Row = TableRun#
Time = TableValue(‘Table 1′,Row-1,#Time)+DELTAt
Duplicate i = 1,9
5-134
T_old[i] = TableValue(‘Table 1′, Row-1,#T[i])
End
“Implicit finite difference formulation”
“Node 1”
h*DELTAx*(T_infi-T[1])+k*DELTAy/2*(300-T[1])/DELTAx+k*DELTAy/2*(T[2]-T[1])/DELTAx+k*DELTAx*(T[4]-
T[1])/DELTAy = rho*c*DELTAx*DELTAy/(2*DELTAt)*(T[1]-T_old[1])
“Node 2”
h*DELTAx*(T_infi-T[2])+k*DELTAy/2*(T[1]-T[2])/DELTAx+k*DELTAy/2*(T[3]-T[2])/DELTAx+k*DELTAx*(T[5]-
T[2])/DELTAy = rho*c*DELTAx*DELTAy/(2*DELTAt)*(T[2]-T_old[2])
“Node 4”
k*DELTAy*(300-T[4])/DELTAx+k*DELTAx*(T[1]-T[4])/DELTAy+k*DELTAy*(T[5]-T[4])/DELTAx+k*DELTAx*(T[7]-
T[4])/DELTAy = rho*c*DELTAx*DELTAy/DELTAt*(T[4]-T_old[4])
“Node 5”
k*DELTAy*(T[4]-T[5])/DELTAx+k*DELTAy*(T[6]-T[5])/DELTAx+k*DELTAx*(T[2]-T[5])/DELTAy+k*DELTAx*(T[8]-
T[5])/DELTAy = rho*c*DELTAx*DELTAy/DELTAt*(T[5]-T_old[5])
“Node 7”
k*DELTAy/2*(300-T[7])/DELTAx+k*DELTAy/2*(T[8]-T[7])/DELTAx+2*k*DELTAx*(T[4]-T[7])/DELTAy =
rho*c*DELTAx*DELTAy/(1*DELTAt)*(T[7]-T_old[7])
“Node 8”
k*DELTAy/2*(T[7]-T[8])/DELTAx+k*DELTAy/2*(T[9]-T[8])/DELTAx+2*k*DELTAx*(T[5]-T[8])/DELTAy =
rho*c*DELTAx*DELTAy/(1*DELTAt)*(T[8]-T_old[8])
“Due to symmetry”
Temperature distribution in the ceramic strip after 12 seconds is as follows
5-135
Special Topic: Controlling the Numerical Error
5-132C The results obtained using a numerical method differ from the exact results obtained analytically because the results
obtained by a numerical method are approximate. The difference between a numerical solution and the exact solution (the
5-133C The discretization error (also called the truncation or formulation error) is due to replacing the derivatives by
differences in each step, or replacing the actual temperature distribution between two adjacent nodes by a straight line
5-135C The Taylor series expansion of the temperature at a specified nodal point m about time ti is
+
+
+=+ 2
2
2),(
2
1
),(
),(),(
t
txT
t
t
txT
ttxTttxT imim
imim
The finite difference formulation of the time derivative at the same nodal point is expressed as
t
TT
t
txTttxT
t
txT i
m
i
mimimim
−
=
−+
+1
),(),(),(
or
t
txT
ttxTttxT im
imim
++ ),(
),(),(
which resembles the Taylor series expansion terminated after the first two terms.
5-136C The Taylor series expansion of the temperature at a specified nodal point m about time ti is
+
+
+=+ 2
2
2),(
2
1
),(
),(),(
t
txT
t
t
txT
ttxTttxT imim
imim
The finite difference formulation of the time derivative at the same nodal point is expressed as
t
TT
t
txTttxT
t
txT i
m
i
mimimim
−
=
−+
+1
),(),(),(
or
t
txT
ttxTttxT im
imim
++ ),(
),(),(
which resembles the Taylor series expansion terminated after the first two terms. Therefore, the 3rd and following terms in
the Taylor series expansion represent the error involved in the finite difference approximation. For a sufficiently small time
step, these terms decay rapidly as the order of derivative increases, and their contributions become smaller and smaller. The
first term neglected in the Taylor series expansion is proportional to
2
)( t
, and thus the local discretization error is also
proportional to
2
)( t
.
The global discretization error is proportional to the step size to t itself since, at the worst case, the accumulated
discretization error after I time steps during a time period
0
t
is
ttttttI == 0
2
0
2)/(
which is proportional to t.
5-137C The round-off error is caused by retaining a limited number of digits during calculations. It depends on the number
PROPRIETARY MATERIAL. © 2020 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
5-139C The round-off error can be reduced by avoiding extremely small mess sizes (smaller than necessary to keep the
5-140C A practical way of checking if the round-off error has been significant in calculations is to repeat the calculations
5-141C A practical way of checking if the discretization error has been significant in calculations is to start the calculations
5-137
Review Problems
5-142 Starting with an energy balance on a volume element,
the steady three-dimensional finite difference equation for a
general interior node in rectangular coordinates for T(x, y, z)
for the case of constant thermal conductivity and uniform heat
generation is to be obtained.
Analysis We consider a volume element of size
zyx
centered about a general interior node (m, n, r) in a region in
which heat is generated at a constant rate of
0
e
and the
thermal conductivity k is variable. Assuming the direction of
heat conduction to be towards the node under consideration at
all surfaces, the energy balance on the volume element can be
expressed as
zyx
element
E
exyz
n+1
•
•
•
•
•
•
n
r+1
r
m+1
m-1
e0
z
x
y
5-138
5-143 Starting with an energy balance on a volume element, the three-dimensional transient explicit finite difference
equation for a general interior node in rectangular coordinates for T(x, y, z, t) for the case of constant thermal conductivity k
and no heat generation is to be obtained.
Analysis We consider a rectangular region in which heat conduction
is significant in the x and y directions. There is no heat generation in
the medium, and the thermal conductivity k of the medium is
constant. Now we divide the x-y-z region into a mesh of nodal points
which are spaced x, y, and z apart in the x, y, and z directions,
•
•
•
z
x
n+1
•
•
m+1
y
5-140
5-146 A plane wall with variable heat generation and variable thermal conductivity is subjected to uniform heat flux
q0
and
convection at the left (node 0) and radiation at the right boundary (node 2). The implicit transient finite difference
formulation of the problem using the energy balance approach method is to be determined.
Assumptions 1 Heat transfer through the wall is given to be
transient, and the thermal conductivity and heat generation
to be variables. 2 Heat transfer is one-dimensional since the
plate is large relative to its thickness. 3 Radiation from the
left surface, and convection from the right surface are
negligible.
Analysis Using the energy balance approach and taking the
direction of all heat transfers to be towards the node under
consideration, the implicit finite difference formulations
become
Left boundary node (node 0):
t
TT
c
x
AxAeTThAAq
x
TT
Ak
ii
p
ii
ii
i
−
=+−++
−+
++
++
+0
1
0
1
0
1
00
1
0
1
1
1
02
)2/()(
Interior node (node 1):
t
TT
xcAxAe
x
TT
Ak
x
TT
Ak
ii
p
i
ii
i
ii
i
−
=+
−
+
−+
+
++
+
++
+1
1
1
1
1
1
1
1
2
1
1
1
1
1
0
1
1)(
Right boundary node (node 2):
t
TT
c
x
AxAeTTA
x
TT
Ak
ii
p
iii
surr
ii
i
−
=++−++
−+
+++
++
+2
1
2
1
2
41
2
41
1
2
1
1
1
22
)2/(])273()273[(
5-147 A pin fin with convection and radiation heat transfer from its tip is considered. The complete finite difference
formulation for the determination of nodal temperatures is to be obtained.
Assumptions 1 Heat transfer through the pin fin is given to be steady and one-dimensional, and the thermal conductivity to
be constant. 2 Convection heat transfer coefficient and emissivity are constant and uniform.
Assumptions 1 Heat transfer through the wall is given to be steady and one–
dimensional, and the thermal conductivity and heat generation to be variable. 2
Convection heat transfer at the right surface is negligible.
Analysis The nodal network consists of 3 nodes, and the base temperature T0 at node
0 is specified. Therefore, there are two unknowns T1 and T2, and we need two
equations to determine them. Using the energy balance approach and taking the
2
Convectio
h, T
•
•
•
0
1
D
x
Convection
x
1
•
•
•
0
2
Tsurr
Radiation
h, T
k(T)
)(xe
0
q